Rotating Vessel- Part II

let us consider two corners A and B. now at corner B, $P_{ 1 }^{ }{ =\quad P_{ 2 }^{ }{ +P_{ 3 }^{ }{ } } }$ .
$P_{ 1 }^{ }{ }$ is the pressure due to the part of the liquid of density $2\sigma$ which is at a height of 3h. $P_{ 2 }^{ }$ is the combined pressure of liquids due to rotation which is situated horizontally. $P_{ 3 }^{ }{ }$ is the pressure due the to part of the liquid of density $\sigma$ which is at a height of h. so, $2\sigma g\left( 3h \right) =\quad \left[ 8\sigma h\overset { 2 }{ } \omega \overset { 2 }{ } +\sigma 33h\overset { 2 }{ } \omega \overset { 2 }{ } \right] \quad +\quad \sigma gh$ . for calculating the pressure due to rotation due to the liquids situated horizontally, distance of the centre of mass of both the liquids from the rotational axis should be taken as the radius of rotation. first calculate the centrifugal force and then divide it by area to get the pressure